5

You are right. There is no such a map as the one you are trying to describe.
Here is a map that actually exists. Let $R$ and $S$ be two noncommutative rings with units, and let $P$ be an $R$-$S$-bimodule. Consider the $S$-$R$-bimodule $Q={}_R\mathrm{Hom}(P,R)$. Then the evaluation is an $R$-$R$-bimodule map
$$
\mathrm{ev}\colon P\otimes_S Q\to R, \...

rt.representation-theory ct.category-theory ra.rings-and-algebras homological-algebra noncommutative-algebra

4

Such examples are a plenty. You are asking about non-existence of an algebra map $A\rightarrow End_AM$. Take $A$ simple, at most countably dimensional, and a simple module ${}_{A}M$. Then $End_AM={\mathbb C}$. Bingo!

4

Doc, you ain't write no evaluation map. If $R$ is commutative, you write the trace map. If $R$ is noncom, god knows what you write. The evaluation map, that is an isomorphism for a finitely generative projective generator and a homomorphism of $R$-$R$-bimodules, in general, is
$$
p \otimes \phi \mapsto \phi (p), \ P \otimes_{End_RP} P^{\vee} \rightarrow R \, ...

rt.representation-theory ct.category-theory ra.rings-and-algebras homological-algebra noncommutative-algebra

4

When the residue characteristic is odd, all supercuspidal representations of GL(2,F) arise from admissible pairs, and via the LLC they correspond two dimensional irreducible continuous representations of Weil group of F.
So I think the word dihedral here should mean that the image of the corresponding Galois representation (in PGL(2, C)) is dihedral.
For ...

3

I take your word that 'dihedral' is Bump's terminology (I don't have the book to hand), but I think that the modern terminology for such supercuspidals is 'tame'. I don't know the first place where it is proven that all supercuspidals of $\operatorname{GL}_2$ are tame in odd residual characteristic; certainly it's proven in Bushnell and Henniart - The local ...

3

Denote by $\mathbb{A}$ the adeles over a number field $F$. An cuspidal automorphic representation $\pi$ of $\text{GL}_2(\mathbb{A})$ now factors by the tensor product theorem as $\pi\cong \prod_v\pi_v$, where $\pi_v$ are smooth irreducible unitary representations of $\text{GL}_2(F_v)$.
The Ramanujan-Petersson conjecture now states that the representations $\...

3

Interesting observation! Unfortunately, this seems to be a coincidence.
The original, algebraically motivated definition of the Hecke algebra gives the quadratic relation as
$$(T-q)(T+1)=0.$$
For $SL_2$, the Hecke algebra has a faithful two dimensional representation. The eigenvalues of $T$ are $q$ and $-1$, and the quadratic relation simply becomes the ...

1

This is a relatively standard abuse of notation, which appears in other areas as well. Some examples:
"Let $X$ be a topological space" rather than "Let $(X, \tau)$ be a topological space"
"Let $V$ be a vector space" rather than "Let $(V, 0, +, -, \cdot)$ be a vector space".
"Let $R$ be a ring" rather than "Let $(R, 0, +, -, \cdot)$ be a ring"
(I welcome ...

1

From the way you phrase your question, I suspect that you misunderstand something. Being supercuspidal is a local condition. If $\pi$ is an automorphic representation of $\mathrm{GL}_2$ over $\mathbb{Q}$, then $\pi_p$ might be supercuspidal for some ramified primes $p$, and non-supercuspidal for some other ramified primes $p$. The Ramanujan conjecture is a ...

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